414 
PROFESSOR G. H. DARWIN ON FIGURES OF 
c= c x + SC 
It is not hard to find an approximate solution c 0 by trial and error, and the correct 
results may then be found thus. 
Consider the case where the two bodies are equal to one another, and put ct = A = 1 . 
The equations then become 
c = 2 4- 
20 
7 , 3 
11 
13 
0 3 + 
3c 3 
— 4--b 
2c 4 ^ e 5 T 
4c G " T 
5 c‘ 5 
C= 1 - 
14 
7 9 
11 
13 
3c 3 
2c 4 2c 5 
1 
1 0 
O 
1 CM 
2c 7 ' 
By trial and error we find c = 2'535, C = + ’557. 
From this we conclude that equilibrium still subsists when the two masses are in 
contact. 
When a — A — 1 , c = 2‘535, we have y = (ct/c ) 2 = g-r^-g-, ai) d /3 = y/(l —y) = 5 -rnn 
Our figures showed that when /3 = w the two masses were nearly in contact, and when 
j 8 = j they crossed. 
This result is, therefore, in accordance with the figures. 
Next pass to the case of an infinitesimal satellite, and suppose a infinitely small 
compared with A and c, and that A = 1 . The equations are 
Then it is obvious that 
Sc 
c n + 
3/3 47 
+ r* + • • • 
and 
and 
c 
1 
+ 
5 
12c 3 ’ 
The solution of the first equation is c = 1'226, and this value of c makes C— — ‘94. 
Hence we conclude that an infinitesimal fluid satellite cannot revolve with its surface 
in contact with its planet. 
C vanishes when c = = 1*89. Hence it appears that the nearest approach of 
the infinitesimal satellite to the planet is 1‘89 mean radii of the planet. The nature 
of the approximation adopted is, however, such that in reality the satellite must 
lie further from the planet than this, perhaps at two radii distance.* The satellite and 
planet of which we here speak are, of course, supposed to revolve as parts of a rigid 
* [See R-oche, ‘ Montpellier, Acad. Sei. Mem.,’ vol. 1, 1S47-1850, p. 243 (added Ocfc. 5, 1887).] 
